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I have trouble proving the next sequence limit:

$\displaystyle\lim_{n\rightarrow\infty}(x_{n}-\sqrt{n})=\frac{1}{2}$

where $x_{n}=\sqrt{n+\sqrt{n-1 ...\sqrt{2+\sqrt{1}}}}.$

I've had a lot of problems; my try is multiplying by the conjugate but the resulting expression continues with $x_{n-1}.$ Also I tried to bound the expression and apply limit in each side which it was failed. By other hand, I thought that I can operate with the expression $x_{n}=\sqrt{n+x_{n-1}}$ and try to get a quadratic equation; solve it and work with a new expression but its usless.

I thank any help to prove this limit.

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By induction,

$$\sqrt n<x_n<\sqrt{n+\sqrt{2n}}.$$

Indeed, $$\sqrt{n+1}<\sqrt{n+\sqrt n}<x_{n+1}=\sqrt{n+x_n}<\sqrt{n+\sqrt{n+\sqrt{2n}}}<\sqrt{n+1+\sqrt{2(n+1)}}.$$

Then $$\sqrt{n+\sqrt n}-\sqrt{n+1}<x_{n+1}-\sqrt{n+1}<\sqrt{n+\sqrt{n+\sqrt{2n}}}-\sqrt{n+1},$$

$$\frac{\sqrt n-1}{\sqrt{n+\sqrt n}+\sqrt{n+1}}<x_{n+1}-\sqrt{n+1}<\frac{\sqrt{n+\sqrt{2n}}-1}{\sqrt{n+\sqrt{n+\sqrt{2n}}}+\sqrt{n+1}}.$$

Both bounds clearly tend to $\dfrac1{1+1}$.

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  • $\begingroup$ Very nice solution! A lot of thanks @YvesDaust! The inductive argument and the monotony of the sequence is the trick behind the proof. $\endgroup$ – Dvd34 Apr 17 '15 at 0:28
  • $\begingroup$ I'd like to add that finding a bracketing with $\sqrt n$ and lower order terms is helpful. If you start from any loose bracketing (say $(0,n)$) and perform a few iterations, the bracketing naturally tightens, hinting you a solution. $\endgroup$ – Yves Daoust Apr 17 '15 at 6:23
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Using that $\sqrt{1+\varepsilon}\approx1+\frac12 \varepsilon$ for $\varepsilon\ll1$.

Now define $a_n\equiv \sqrt{n-1 ...\sqrt{2+\sqrt{1}}}\ll n$ for big $n\gg 1$ you get $a_n\ll n$ then:

$$\displaystyle\lim_{n\rightarrow\infty}(x_{n}-\sqrt{n})=\displaystyle\lim_{n\rightarrow\infty}(\sqrt{n+a_n}-\sqrt{n}) \approx \displaystyle\lim_{n\rightarrow\infty}\left(\sqrt{n}\left(1+\frac12\frac{ a_n}{n}\right)-\sqrt{n}\right)= \frac12\displaystyle\lim_{n\rightarrow\infty}\left(\frac{a_n}{\sqrt{n}}\right)$$

Then you need to show for $n\gg1$ that $$a_n=\sqrt n$$

Maybe: $$a_n^2\approx(n-1)+\frac12\frac{n-2}{n-1}+\frac14\frac{n-3}{n-1}+... \approx (n-1)+\frac12+\frac14+...=(n-1)+1 =n$$

Because $$\sum_{k=0}^\infty\frac{1}{2^k}=\frac{1}{1-\frac12}=2$$

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this might be a direction... $$ \frac{x_n}{\sqrt{n}} = \sqrt{1+\sqrt{\frac{n-1}{n^2}+\sqrt\frac{n-2}{n^4}}\cdots} \to 1+ \frac1{2\sqrt{n}} $$

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  • $\begingroup$ The $\rightarrow$ here isn't necessarily true, since although each error term $\rightarrow 0$, there are increasingly many of them. $\endgroup$ – Christopher Apr 16 '15 at 11:02
  • $\begingroup$ you are right, but i thought that, since this manipulation gives the required result, there may be a more detailed analysis which could put the inference on a sounder basis. $\endgroup$ – David Holden Apr 16 '15 at 11:33

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